YES * Step 1: UnsatRules YES + Considered Problem: Rules: 0. start(A,B,C,D,E,F) -> stop(A,B,C,D,E,F) [0 >= A && B = C && D = A && E = F] (?,1) 1. start(A,B,C,D,E,F) -> lbl6(A,B,C,D,E,F) [A >= 1 && A >= C && B = C && D = A && E = F] (?,1) 2. start(A,B,C,D,E,F) -> lbl121(A,B,C,D,B + -1*D,F) [A >= 1 && C >= 1 + A && B = C && D = A && E = F] (?,1) 3. lbl6(A,B,C,D,E,F) -> stop(A,B,C,D,E,F) [A >= 1 && A >= C && E = F && D = A && B = C] (?,1) 4. lbl111(A,B,C,D,E,F) -> cut(A,B,C,D,E,F) [C >= 1 + A && A >= 2 && E = 0 && D = A && B = C] (?,1) 5. lbl111(A,B,C,D,E,F) -> lbl111(A,B,C,D,-1 + E,F) [E >= 1 && A >= 1 + E && E >= 0 && C >= 1 + A + E && A >= 2 + E && D = A && B = C] (?,1) 6. lbl111(A,B,C,D,E,F) -> lbl121(A,B,C,D,-1*D + E,F) [E >= 1 && E >= A && E >= 0 && C >= 1 + A + E && A >= 2 + E && D = A && B = C] (?,1) 7. lbl121(A,B,C,D,E,F) -> cut(A,B,C,D,E,F) [C >= 1 + A && A >= 1 && C >= A && E = 0 && D = A && B = C] (?,1) 8. lbl121(A,B,C,D,E,F) -> lbl111(A,B,C,D,-1 + E,F) [E >= 1 && A >= 1 + E && C >= 1 + A && A >= 1 && E >= 0 && C >= A + E && D = A && B = C] (?,1) 9. lbl121(A,B,C,D,E,F) -> lbl121(A,B,C,D,-1*D + E,F) [E >= 1 && E >= A && C >= 1 + A && A >= 1 && E >= 0 && C >= A + E && D = A && B = C] (?,1) 10. cut(A,B,C,D,E,F) -> stop(A,B,C,D,E,F) [A >= 1 && C >= 1 + A && E = 0 && D = A && B = C] (?,1) 11. start0(A,B,C,D,E,F) -> start(A,C,C,A,F,F) True (1,1) Signature: {(cut,6);(lbl111,6);(lbl121,6);(lbl6,6);(start,6);(start0,6);(stop,6)} Flow Graph: [0->{},1->{3},2->{7,8,9},3->{},4->{10},5->{4,5,6},6->{7,8,9},7->{10},8->{4,5,6},9->{7,8,9},10->{},11->{0,1 ,2}] + Applied Processor: UnsatRules + Details: Following transitions have unsatisfiable constraints and are removed: [6] * Step 2: UnsatPaths YES + Considered Problem: Rules: 0. start(A,B,C,D,E,F) -> stop(A,B,C,D,E,F) [0 >= A && B = C && D = A && E = F] (?,1) 1. start(A,B,C,D,E,F) -> lbl6(A,B,C,D,E,F) [A >= 1 && A >= C && B = C && D = A && E = F] (?,1) 2. start(A,B,C,D,E,F) -> lbl121(A,B,C,D,B + -1*D,F) [A >= 1 && C >= 1 + A && B = C && D = A && E = F] (?,1) 3. lbl6(A,B,C,D,E,F) -> stop(A,B,C,D,E,F) [A >= 1 && A >= C && E = F && D = A && B = C] (?,1) 4. lbl111(A,B,C,D,E,F) -> cut(A,B,C,D,E,F) [C >= 1 + A && A >= 2 && E = 0 && D = A && B = C] (?,1) 5. lbl111(A,B,C,D,E,F) -> lbl111(A,B,C,D,-1 + E,F) [E >= 1 && A >= 1 + E && E >= 0 && C >= 1 + A + E && A >= 2 + E && D = A && B = C] (?,1) 7. lbl121(A,B,C,D,E,F) -> cut(A,B,C,D,E,F) [C >= 1 + A && A >= 1 && C >= A && E = 0 && D = A && B = C] (?,1) 8. lbl121(A,B,C,D,E,F) -> lbl111(A,B,C,D,-1 + E,F) [E >= 1 && A >= 1 + E && C >= 1 + A && A >= 1 && E >= 0 && C >= A + E && D = A && B = C] (?,1) 9. lbl121(A,B,C,D,E,F) -> lbl121(A,B,C,D,-1*D + E,F) [E >= 1 && E >= A && C >= 1 + A && A >= 1 && E >= 0 && C >= A + E && D = A && B = C] (?,1) 10. cut(A,B,C,D,E,F) -> stop(A,B,C,D,E,F) [A >= 1 && C >= 1 + A && E = 0 && D = A && B = C] (?,1) 11. start0(A,B,C,D,E,F) -> start(A,C,C,A,F,F) True (1,1) Signature: {(cut,6);(lbl111,6);(lbl121,6);(lbl6,6);(start,6);(start0,6);(stop,6)} Flow Graph: [0->{},1->{3},2->{7,8,9},3->{},4->{10},5->{4,5},7->{10},8->{4,5},9->{7,8,9},10->{},11->{0,1,2}] + Applied Processor: UnsatPaths + Details: We remove following edges from the transition graph: [(2,7)] * Step 3: FromIts YES + Considered Problem: Rules: 0. start(A,B,C,D,E,F) -> stop(A,B,C,D,E,F) [0 >= A && B = C && D = A && E = F] (?,1) 1. start(A,B,C,D,E,F) -> lbl6(A,B,C,D,E,F) [A >= 1 && A >= C && B = C && D = A && E = F] (?,1) 2. start(A,B,C,D,E,F) -> lbl121(A,B,C,D,B + -1*D,F) [A >= 1 && C >= 1 + A && B = C && D = A && E = F] (?,1) 3. lbl6(A,B,C,D,E,F) -> stop(A,B,C,D,E,F) [A >= 1 && A >= C && E = F && D = A && B = C] (?,1) 4. lbl111(A,B,C,D,E,F) -> cut(A,B,C,D,E,F) [C >= 1 + A && A >= 2 && E = 0 && D = A && B = C] (?,1) 5. lbl111(A,B,C,D,E,F) -> lbl111(A,B,C,D,-1 + E,F) [E >= 1 && A >= 1 + E && E >= 0 && C >= 1 + A + E && A >= 2 + E && D = A && B = C] (?,1) 7. lbl121(A,B,C,D,E,F) -> cut(A,B,C,D,E,F) [C >= 1 + A && A >= 1 && C >= A && E = 0 && D = A && B = C] (?,1) 8. lbl121(A,B,C,D,E,F) -> lbl111(A,B,C,D,-1 + E,F) [E >= 1 && A >= 1 + E && C >= 1 + A && A >= 1 && E >= 0 && C >= A + E && D = A && B = C] (?,1) 9. lbl121(A,B,C,D,E,F) -> lbl121(A,B,C,D,-1*D + E,F) [E >= 1 && E >= A && C >= 1 + A && A >= 1 && E >= 0 && C >= A + E && D = A && B = C] (?,1) 10. cut(A,B,C,D,E,F) -> stop(A,B,C,D,E,F) [A >= 1 && C >= 1 + A && E = 0 && D = A && B = C] (?,1) 11. start0(A,B,C,D,E,F) -> start(A,C,C,A,F,F) True (1,1) Signature: {(cut,6);(lbl111,6);(lbl121,6);(lbl6,6);(start,6);(start0,6);(stop,6)} Flow Graph: [0->{},1->{3},2->{8,9},3->{},4->{10},5->{4,5},7->{10},8->{4,5},9->{7,8,9},10->{},11->{0,1,2}] + Applied Processor: FromIts + Details: () * Step 4: Decompose YES + Considered Problem: Rules: start(A,B,C,D,E,F) -> stop(A,B,C,D,E,F) [0 >= A && B = C && D = A && E = F] start(A,B,C,D,E,F) -> lbl6(A,B,C,D,E,F) [A >= 1 && A >= C && B = C && D = A && E = F] start(A,B,C,D,E,F) -> lbl121(A,B,C,D,B + -1*D,F) [A >= 1 && C >= 1 + A && B = C && D = A && E = F] lbl6(A,B,C,D,E,F) -> stop(A,B,C,D,E,F) [A >= 1 && A >= C && E = F && D = A && B = C] lbl111(A,B,C,D,E,F) -> cut(A,B,C,D,E,F) [C >= 1 + A && A >= 2 && E = 0 && D = A && B = C] lbl111(A,B,C,D,E,F) -> lbl111(A,B,C,D,-1 + E,F) [E >= 1 && A >= 1 + E && E >= 0 && C >= 1 + A + E && A >= 2 + E && D = A && B = C] lbl121(A,B,C,D,E,F) -> cut(A,B,C,D,E,F) [C >= 1 + A && A >= 1 && C >= A && E = 0 && D = A && B = C] lbl121(A,B,C,D,E,F) -> lbl111(A,B,C,D,-1 + E,F) [E >= 1 && A >= 1 + E && C >= 1 + A && A >= 1 && E >= 0 && C >= A + E && D = A && B = C] lbl121(A,B,C,D,E,F) -> lbl121(A,B,C,D,-1*D + E,F) [E >= 1 && E >= A && C >= 1 + A && A >= 1 && E >= 0 && C >= A + E && D = A && B = C] cut(A,B,C,D,E,F) -> stop(A,B,C,D,E,F) [A >= 1 && C >= 1 + A && E = 0 && D = A && B = C] start0(A,B,C,D,E,F) -> start(A,C,C,A,F,F) True Signature: {(cut,6);(lbl111,6);(lbl121,6);(lbl6,6);(start,6);(start0,6);(stop,6)} Rule Graph: [0->{},1->{3},2->{8,9},3->{},4->{10},5->{4,5},7->{10},8->{4,5},9->{7,8,9},10->{},11->{0,1,2}] + Applied Processor: Decompose NoGreedy + Details: We construct a looptree: P: [0,1,2,3,4,5,7,8,9,10,11] | +- p:[9] c: [9] | `- p:[5] c: [5] * Step 5: CloseWith YES + Considered Problem: (Rules: start(A,B,C,D,E,F) -> stop(A,B,C,D,E,F) [0 >= A && B = C && D = A && E = F] start(A,B,C,D,E,F) -> lbl6(A,B,C,D,E,F) [A >= 1 && A >= C && B = C && D = A && E = F] start(A,B,C,D,E,F) -> lbl121(A,B,C,D,B + -1*D,F) [A >= 1 && C >= 1 + A && B = C && D = A && E = F] lbl6(A,B,C,D,E,F) -> stop(A,B,C,D,E,F) [A >= 1 && A >= C && E = F && D = A && B = C] lbl111(A,B,C,D,E,F) -> cut(A,B,C,D,E,F) [C >= 1 + A && A >= 2 && E = 0 && D = A && B = C] lbl111(A,B,C,D,E,F) -> lbl111(A,B,C,D,-1 + E,F) [E >= 1 && A >= 1 + E && E >= 0 && C >= 1 + A + E && A >= 2 + E && D = A && B = C] lbl121(A,B,C,D,E,F) -> cut(A,B,C,D,E,F) [C >= 1 + A && A >= 1 && C >= A && E = 0 && D = A && B = C] lbl121(A,B,C,D,E,F) -> lbl111(A,B,C,D,-1 + E,F) [E >= 1 && A >= 1 + E && C >= 1 + A && A >= 1 && E >= 0 && C >= A + E && D = A && B = C] lbl121(A,B,C,D,E,F) -> lbl121(A,B,C,D,-1*D + E,F) [E >= 1 && E >= A && C >= 1 + A && A >= 1 && E >= 0 && C >= A + E && D = A && B = C] cut(A,B,C,D,E,F) -> stop(A,B,C,D,E,F) [A >= 1 && C >= 1 + A && E = 0 && D = A && B = C] start0(A,B,C,D,E,F) -> start(A,C,C,A,F,F) True Signature: {(cut,6);(lbl111,6);(lbl121,6);(lbl6,6);(start,6);(start0,6);(stop,6)} Rule Graph: [0->{},1->{3},2->{8,9},3->{},4->{10},5->{4,5},7->{10},8->{4,5},9->{7,8,9},10->{},11->{0,1,2}] ,We construct a looptree: P: [0,1,2,3,4,5,7,8,9,10,11] | +- p:[9] c: [9] | `- p:[5] c: [5]) + Applied Processor: CloseWith True + Details: () YES